The Normal Distribution — A-Level Mathematics Stats Study Guide

For: A-Level Mathematics candidates sitting Paper 5 (Probability & Statistics 1).

Covers: All core normal distribution skills required for Paper 5, including properties of the normal curve, standardisation, z-table calculations, binomial approximation with continuity correction, and inverse normal problem solving.

You should already know: Basic probability, summation, integration (Pure 1 calculus).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is The Normal Distribution?

The normal distribution, also called the Gaussian distribution or bell curve, is a continuous probability distribution that describes symmetric, unskewed data where values cluster around a central mean, with frequencies tapering off equally in both directions away from the mean. It is one of the most widely used distributions in statistics, as it models many real-world phenomena including human heights, standardized test scores, and measurement errors.

For a random variable X following a normal distribution, we use the notation $X\sim N(\mu ,{\sigma }^2)$, where $\mu$ is the population mean and ${\sigma }^2$ is the population variance. This topic makes up 15-20% of total marks on A-Level Mathematics Paper 5, and is often combined with binomial probability or conditional probability questions in longer 6-8 mark problems.

2. Properties of $N(\mu ,{\sigma }^2)$

The normal distribution has a set of fixed, predictable properties that form the basis of all related calculations:

1. Perfect symmetry: The curve is symmetric around the mean $\mu$, so the mean, median, and mode of the distribution are all equal. This means $P(X<\mu )=P(X>\mu )=0.5$, and $P(X<\mu -a)=P(X>\mu +a)$ for any positive value of $a$.
2. Total area = 1: As a probability density function (PDF), the total area under the entire normal curve equals 1, representing the sum of all possible probabilities for the variable.
3. Spread determined by standard deviation: A larger value of $\sigma$ (standard deviation) produces a wider, flatter curve, while a smaller $\sigma$ produces a narrower, taller curve, with no change to the center of the distribution.
4. Empirical (68-95-99.7) rule: Roughly 68% of all observed values fall within $\mu \pm \sigma$, 95% within $\mu \pm 2\sigma$, and 99.7% within $\mu \pm 3\sigma$. This rule is useful for quick sanity checks of your calculations.
5. PDF form: The full probability density function is $f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}$: you do not need to memorize this, but you should recognize that integrating this function between two bounds gives the probability of X falling in that interval.

Worked example: The mass of loaves of bread produced by a bakery is $X\sim N(800,36)$, so $\mu =800g$ and $\sigma =6g$. Using the empirical rule, 95% of loaves will have a mass between $800-2\ast 6=788g$ and $800+2\ast 6=812g$. By symmetry, the probability that a loaf is lighter than 794g is equal to the probability it is heavier than 806g.

3. Standardisation $Z=\frac{X-\mu }{\sigma }$

Since there are infinitely many possible combinations of $\mu$ and ${\sigma }^2$ for normal distributions, we cannot precompute probability tables for every possible distribution. Instead, we convert any normal distribution to the standard normal distribution Z, which has fixed parameters $\mu =0$ and ${\sigma }^2=1$ (so $\sigma =1$), written $Z\sim N(0,1)$.

The standardisation formula converts a raw score X from any normal distribution to a z-score, which measures how many standard deviations X is above or below the mean: $$Z=\frac{X-\mu }{\sigma }$$ Where:

• $X$ = raw score from the original $N(\mu ,{\sigma }^2)$ distribution
• $\mu$ = mean of the original distribution
• $\sigma$ = standard deviation of the original distribution
• $Z$ = standardised z-score, following $N(0,1)$

A positive z-score means X is above the mean, a negative z-score means X is below the mean, and a z-score of 0 means X equals the mean.

Worked example: The IQ scores of a group of students are $X\sim N(110,225)$, so $\sigma =15$. Calculate the z-score for a student with an IQ of 137: $$Z=\frac{137-110}{15}=\frac{27}{15}=1.8$$ This means the student’s IQ is 1.8 standard deviations above the group mean. For a student with an IQ of 92, $Z=\frac{92-110}{15}=-1.2$, so their score is 1.2 standard deviations below the mean.

4. Using normal tables / z-tables

A-Level provides standard normal (z) tables in exams that give the cumulative probability $P(Z<z)$ for positive values of z, to 2 decimal places. The leftmost column of the table gives the first two digits of the z-score (e.g. 0.2, 1.5, 2.0), and the top row gives the second decimal place (0.00, 0.01, ..., 0.09).

Use these symmetry rules to handle negative z-scores and upper-tail probabilities:

1. $P(Z<-a)=P(Z>a)=1-P(Z<a)$ for any positive $a$
2. $P(Z>a)=1-P(Z<a)$ for any $a$
3. $P(a<Z<b)=P(Z<b)-P(Z<a)$ for any $a<b$

Exam tip: Always use the table values provided in the exam, not calculator-generated values, as mark schemes are written to match the precision of official A-Level tables.

Worked examples:

1. Find $P(Z<1.36)$: Look for the row labeled 1.3 and column labeled 0.06: the value is 0.9131, so $P(Z<1.36)=0.9131$.
2. Find $P(Z>0.72)$: $1-P(Z<0.72)=1-0.7642=0.2358$.
3. Find $P(-0.45<Z<1.12)$: $P(Z<1.12)-P(Z<-0.45)=0.8686-(1-0.6736)=0.8686-0.3264=0.5422$.

5. Approximating binomial with continuity correction

The binomial distribution $X\sim B(n,p)$ describes discrete count data, but when $n$ is large and $p$ is not close to 0 or 1, it can be approximated by a normal distribution. For A-Level exams, the approximation is valid if $np>5$ and $nq>5$, where $q=1-p$.

The parameters of the normal approximation are: $$\mu =np,{\sigma }^2=npq$$ So we use $Y\sim N(np,npq)$ to approximate $X\sim B(n,p)$.

Continuity correction

Since the binomial distribution is discrete and the normal distribution is continuous, we apply a continuity correction to adjust for the difference between discrete and continuous variables:

• For discrete $X=k$, the equivalent continuous interval is $k-0.5$ to $k+0.5$
• $P(X\le k)\approx P(Y\le k+0.5)$
• $P(X\ge k)\approx P(Y\ge k-0.5)$
• $P(X<k)=P(X\le k-1)\approx P(Y\le k-0.5)$
• $P(X>k)=P(X\ge k+1)\approx P(Y\ge k+0.5)$

Exam tip: Examiners deduct 1-2 marks for missing continuity correction in binomial-normal approximation questions, so never skip this step.

Worked example: A fair 6-sided die is rolled 120 times, and X is the number of times a 6 is rolled. Use a normal approximation to find $P(X\ge 25)$.

1. First, $X\sim B(120,1/6)$. Check validity: $np=120\ast 1/6=20>5$, $nq=120\ast 5/6=100>5$, so approximation is valid.
2. Calculate normal parameters: $\mu =20$, ${\sigma }^2=120\ast 1/6\ast 5/6=100/6\approx 16.67$, so $\sigma \approx 4.08$.
3. Apply continuity correction: $P(X\ge 25)\approx P(Y\ge 24.5)$ where $Y\sim N(20,16.67)$.
4. Standardise: $Z=\frac{24.5-20}{4.08}\approx 1.10$.
5. Calculate probability: $P(Z\ge 1.10)=1-0.8643=0.1357$.

6. Inverse normal — finding $x$ given probability

In many exam questions, you are given a probability and asked to find the corresponding raw score $x$ or z-score $z$. This is called an inverse normal problem.

Follow these steps to solve inverse normal questions:

1. Rewrite the given probability in the form $P(X<x)=p$, using symmetry to adjust for upper-tail probabilities if needed.
2. Look up the z-score that corresponds to cumulative probability $p$, using the inverse normal table or reverse lookup in the standard normal table.
3. Rearrange the standardisation formula to solve for $x$: $$x=\mu +z\sigma$$

Exam tip: If $p<0.5$, the corresponding z-score is negative, and if $p>0.5$, z is positive. Use this as a quick check to avoid sign errors.

Worked example: The heights of 16-year-old boys are normally distributed with mean 175 cm and standard deviation 6 cm. The tallest 15% of boys qualify for a regional basketball team. Find the minimum height required to qualify.

1. We need $x$ such that $P(X>x)=0.15$, so $P(X<x)=0.85$.
2. Look up the z-score for $p=0.85$: the closest table value is $z=1.04$, since $P(Z<1.04)=0.8508$.
3. Calculate $x$: $x=175+(1.04\ast 6)=175+6.24=181.24$ cm, so the minimum height is 181 cm (3 significant figures, per A-Level conventions).

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using $\sigma$ instead of ${\sigma }^2$ in normal distribution notation, e.g. writing $N(100,15)$ instead of $N(100,225)$ when $\sigma =15$. Why students do it: Confusing variance and standard deviation in the $N(\mu ,{\sigma }^2)$ notation. Correct move: Always remember the second parameter of the normal distribution is variance, so square the standard deviation when writing the distribution, and take the square root of the second parameter when calculating z-scores.
• Wrong move: Skipping continuity correction for binomial-normal approximations. Why students do it: Forgetting that the binomial distribution is discrete while the normal distribution is continuous. Correct move: Any time you use a normal approximation for a binomial variable, adjust the boundary value by ±0.5 according to the inequality direction before standardising.
• Wrong move: Using calculator-generated z-values instead of official A-Level table values. Why students do it: Relying on their calculator’s built-in normal function instead of the provided tables. Correct move: Always use z-values from the official A-Level tables, as these are the only values accepted in mark schemes. For inverse normal questions, use the closest table value to the given probability.
• Wrong move: Using a positive z-score for probabilities less than 0.5 in inverse normal problems. Why students do it: Forgetting the symmetry of the normal curve. Correct move: If $P(X<x)<0.5$, x is below the mean so z must be negative, and vice versa. Always cross-check your z sign against the mean before calculating x.
• Wrong move: Calculating upper-tail probabilities directly from the z-table, e.g. writing $P(Z>1.2)=0.8849$ instead of $1-0.8849=0.1151$. Why students do it: Misreading the table label, which only gives cumulative probability below a given z-score. Correct move: Explicitly label the area you are calculating, and subtract from 1 for all upper-tail or negative z probabilities.

8. Practice Questions (Paper 5 Style)

Question 1

The monthly electricity bill for households in a town is normally distributed with mean $120andstandarddeviation$22. (a) Find the probability that a randomly selected household has a monthly bill less than $90. (b) Find the probability that a randomly selected household has a monthly bill between $100and$150.

Solution 1

$X\sim N(120,22^2)=N(120,484)$ (a) $P(X<90)$: Standardise: $Z=\frac{90-120}{22}\approx -1.36$ $P(Z<-1.36)=1-P(Z<1.36)=1-0.9131=0.0869$

(b) $P(100<X<150)=P(X<150)-P(X<100)$ Standardise 150: $Z=\frac{150-120}{22}\approx 1.36$, so $P(Z<1.36)=0.9131$ Standardise 100: $Z=\frac{100-120}{22}\approx -0.91$, so $P(Z<-0.91)=1-0.8186=0.1814$ Probability = $0.9131-0.1814=0.7317$

Question 2

A biased coin has a 0.4 probability of landing on heads. The coin is tossed 200 times, and X is the number of heads obtained. Use a normal approximation to find the probability that the number of heads is between 70 and 90 inclusive.

Solution 2

$X\sim B(200,0.4)$ Check validity: $np=200\ast 0.4=80>5$, $nq=200\ast 0.6=120>5$, so approximation is valid. Normal parameters: $\mu =80$, ${\sigma }^2=200\ast 0.4\ast 0.6=48$, so $\sigma \approx 6.93$ Apply continuity correction: $P(70\le X\le 90)\approx P(69.5<Y<90.5)$ where $Y\sim N(80,48)$ Standardise 69.5: $Z=\frac{69.5-80}{6.93}\approx -1.52$, so $P(Z<-1.52)=1-0.9357=0.0643$ Standardise 90.5: $Z=\frac{90.5-80}{6.93}\approx 1.52$, so $P(Z<1.52)=0.9357$ Probability = $0.9357-0.0643=0.8714$

Question 3

The time taken for a baker to bake a loaf of bread is normally distributed with mean 45 minutes. It is found that 3% of loaves take longer than 52 minutes to bake. Find the standard deviation of the baking time, correct to 2 decimal places.

Solution 3

$X\sim N(45,{\sigma }^2)$ Given $P(X>52)=0.03$, so $P(X<52)=0.97$ Find z-score for $p=0.97$: closest table value is $z=1.88$, since $P(Z<1.88)=0.9699\approx 0.97$ Rearrange standard formula: $z=\frac{x-\mu }{\sigma }\to 1.88=\frac{52-45}{\sigma }$ Solve for $\sigma$: $\sigma =\frac{7}{1.88}\approx 3.72$ minutes.

9. Quick Reference Cheatsheet

10. What's Next

The normal distribution is a foundational topic for both remaining A-Level Mathematics content and further quantitative study. In Paper 6 (Probability & Statistics 2), you will use normal distributions to construct confidence intervals, conduct hypothesis tests, and approximate Poisson distributions, so mastering the skills in this guide is non-negotiable for success in later A Level Maths components. It also frequently appears in cross-topic Paper 5 questions combined with permutations, combinations, and conditional probability, so practice integrating these skills with other topics to score full marks on longer 6-8 mark exam questions.

If you struggle with any of the concepts, practice questions, or step-by-step calculations in this guide, you can ask Ollie, our AI tutor, for personalized explanations, extra practice problems, or feedback on your work at any time. You can also find more study guides, past paper walkthroughs, and topic quizzes on the OwlsPrep homepage to help you prepare for your A-Level Mathematics Paper 5 exam.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.